Friday, April 28, 2017

The Thermodynamics of the Proton Spin

In Intermediate Physics for Medicine and Biology, Russ Hobbie and I introduce a lot of statistical mechanics and thermodynamics. For instance, in Chapter 3 we describe the Boltzmann factor and the heat capacity, and in Chapter 18 we analyze magnetic resonance imaging by considering the magnetization of the two-state spin system using statistical mechanics. Perhaps we can do a little more.

First, let’s calculate the average energy of a spin-1/2 proton in a magnetic field B. The proton has two states: up and down. Up has the lower energy Eup = -μB, and down has the higher energy Edown = +μB, where μ is the proton’s magnetic moment. Using the Boltzmann factor, the probability of having spin up is Pup = C eμB/kT, and of spin down is Pdown = C e-μB/kT, where T is the absolute temperature and k is Boltzmann’s constant. The spin must be in one of these two states, so Pup + Pdown = 1, or C = 1/(eμB/kT + e-μB/kT). The average energy,E⟩, is PupEup + PdownEdown, or

An equation giving the average energy of spins in a magnetic field B at a temperature T.
The total energy E of the spin system is just the average energy times the number of spins, E = NE.

Equations are not just things you plug numbers into to get other numbers. Equations tell a physical story. So, whenever I teach the two-state spin system I stress the story, which becomes clearer if we examine the limiting cases of this equation. At high temperatures (μB much less than kT), the argument of the hyperbolic tangent is small, we can use a Taylor expansion for the exponentials, and the average energy is –μ2B2/kT. This is the limit of interest for magnetic resonance imaging, when the average energy increases as the square of the magnetic field. At low temperatures (μB much greater than kT), the argument of the hyperbolic tangent is large, tanh goes to one, and the average energy is –μB. All the spins are in the spin up ground state.

Next, let’s calculate the heat capacity, C = dE/dT. The derivative of the hyperbolic tangent is the hyperbolic secant squared, so

An equation for the heat capacity of spins in a magnetic field B at temperature T.
The leading factor is the number of molecules time the Boltzmann constant, which is equal to the number of moles times the gas constant. At high temperatures, C goes to zero because of the leading factor of 1/T2. Physically, this result arises because in this case the spins are approximately half spin up and half spin down, so the average energy is about zero, and making the system even hotter won’t change the situation. You typically see this type of behavior in systems that have an upper energy level (as opposed to, say, a system like the harmonic oscillator that has energy levels at increasing energies without bound). At low temperatures, C also goes to zero because the secant goes to zero at large argument. This result arises because the spin down state freezes out: if the system is cold enough no spins can reach the spin down state so the average energy is simply the energy of the spin up ground state.

The heat capacity going to zero as the temperature goes to zero is one way of stating the third law of thermodynamics. Russ and I discuss the first and second laws of thermodynamics in IPMB, but not the third. This is mainly because life occurs at warm temperatures, so the behavior as T approaches absolute zero does not have much biological significance. But although little biology happens around absolute zero, much physics does. To learn more about the world at low temperatures, I recommend the book The Quest for Absolute Zero, by Kurt Mendelssohn. Fascinating reading.

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